Question

For each operation \[ * \] defined below, determine whether \[ * \] is binary, commutative or associative.
(i)On \[{\bf{Z}}\] defined \[a * b = a - b\]
(ii)On \[{\bf{Q}}\] defined \[a * b = ab + 1\]
(iii)On \[{\bf{Q}}\] defined \[a * b = \dfrac{{ab}}{2}\]
(iv)On \[{{\bf{Z}}^ + }\] defined \[a * b = {2^{ab}}\]
(v)On \[{{\bf{Z}}^ + }\] defined \[a * b = {a^b}\]
(vi)On \[{\bf{R}} - \left\{ { - 1} \right\}\] defined \[a * b = \dfrac{a}{{b + 1}}\]

Answer 1

Hint: Here, we have to find the operations whether the operation is binary, commutative, or associative. The binary operation can be defined as an operation * which is performed on a set A. The commutative property states that the numbers on which we operate can be moved or swapped from their position without making any difference to the answer. Associative Property states that when three or more numbers are added, the sum is the same regardless of the grouping of the addends.

Complete step-by-step answer:
(i) On \[{\bf{Z}}\], define \[a * b = a - b\]
Here \[a \in {\bf{Z}}\] and \[b \in {\bf{Z}}\] i.e. ,\[a\] and \[b\] are integers.
Let a=2,b=5 \[a = 2\] and \[b = 5\].
Substituting in \[a * b = a - b\], we get
\[ \Rightarrow 2 * 5 = 2 - 5 = - 3\]
As, \[ - 3\] is an integer i.e., \[ - 3 \in {\bf{Z}}\].
Hence,\[*\] is a binary operation.
We know that Subtraction is a binary operation on integers.
Now, we will check for commutative.
∗ is commutative if, \[a * b = b * a\] .
\[\begin{array}{l}a * b = b * a\\ \Rightarrow a - b = b - a\end{array}\]
Substituting \[a = 2,b = 5\] in above equation, we get
\[\begin{array}{l} \Rightarrow 2 - 5 = 5 - 2\\ \Rightarrow - 3 \ne 3\end{array}\]
Hence,∗ is not commutative.
Now, we have to check for associative
∗ is associative if \[(a * b) * c = a * (b * c)\]
\[\begin{array}{l}a * (b * c) = (a * b) * c\\ \Rightarrow a - \left( {b - c} \right) = \left( {a - b} \right) - c\end{array}\]
Let \[a = 2;b = 5;c = 3\]
Substituting \[a = 2;b = 5;c = 3\] in the above equation, we get
\[\begin{array}{l} \Rightarrow 2 - \left( {5 - 3} \right) = \left( {2 - 5} \right) - 3\\ \Rightarrow 2 - 2 = - 3 - 3\end{array}\]
\[ \Rightarrow 0 \ne - 6\]
Hence,∗ is not associative.
(ii)On \[{\bf{Q}}\],define \[a * b = ab + 1\]
Now, we have to check for commutative.
∗ is commutative if, \[a * b = b * a\].
\[a * b = ab + 1\];
\[b * a = ba + 1 = ab + 1\];
Since \[a * b = b * a\] for all \[a,b \in {\bf{Q}}\].
∗ is commutative.
Now, we have to check for associative
∗ is associative if \[(a * b) * c = a * (b * c)\]
\[\left( {ab + 1} \right) * c = a * \left( {bc + 1} \right) \Rightarrow \left( {ab + 1} \right)c + 1 = a\left( {bc + 1} \right) + 1\]
\[ \Rightarrow abc + c + 1 \ne abc + a + 1\]
Since \[(a * b) * c \ne a * (b * c)\]
∗ is not an associative binary operation.

(iii)On \[{\bf{Q}}\] , define \[a * b = \dfrac{{ab}}{2}\]
Now, we have to check for commutative.
∗ is commutative if \[a * b = b * a\]
\[ \Rightarrow a * b = \dfrac{{ab}}{2}\]
​\[ \Rightarrow b * a = \dfrac{{ba}}{2} = \dfrac{{ab}}{2}\]
​\[a * b = b * a\forall a,b \in {\bf{Q}}\]
∗ is commutative.
Now, we have to check for associations.
∗ is associative if \[(a * b) * c = a * (b * c)\]
\[ \Rightarrow (a * b) * c = \dfrac{{ab}}{2} * c = \dfrac{{\dfrac{{ab}}{2}c}}{2} = \dfrac{{abc}}{4}\]
\[ \Rightarrow a * (b * c) = a * \dfrac{{bc}}{2} = \dfrac{{a\dfrac{{bc}}{2}}}{2} = \dfrac{{abc}}{4}\] ​
Since \[(a * b) * c = a * (b * c)\forall a,b,c \in {\bf{Q}}\]
∗ is an associative binary operation.

(iv) On \[{{\bf{Z}}^ + }\], define if \[a * b = b * a\]
Now, we have to check for commutative.
∗ is commutative if \[a * b = b * a\]
\[ \Rightarrow a * b = {2^{ab}}\]
\[ \Rightarrow b * a = {2^{ba}} = {2^{ab}}\]
Since \[a * b = b * a\forall a,b \in {{\bf{Z}}^ + }\]
∗ is commutative.
Now, we have to check for associations.
∗ is associative if \[(a * b) * c = a * (b * c)\]
\[ \Rightarrow (a * b) * c = {2^{ab}} * c = {2^{{2^{ab}}c}}\]
\[ \Rightarrow a * \left( {b * c} \right) = a * {2^{bc}} = {2^{a{2^{bc}}}}\]
Since \[(a * b) * c \ne a * (b * c)\]
∗ is not an associative binary operation.

(v) On \[{{\bf{Z}}^ + }\] define \[a * b = {a^b}\]
Now, we have to check for commutative.
∗ is commutative if \[a * b = b * a\]
\[ \Rightarrow a * b = {a^b}\]
\[ \Rightarrow b * a = {b^a}\]
\[ \Rightarrow a * b \ne b * a\]
∗ is not commutative.
Now, we have to check for associations.
∗ is associative if \[(a * b) * c = a * (b * c)\]
\[ \Rightarrow (a * b) * c = {a^b} * c = {\left( {{a^b}} \right)^c}\]
\[ \Rightarrow a * (b * c) = a * {b^c} = {(a)^{{b^c}}}\]
Since \[(a * b) * c \ne a * (b * c)\]
∗ is not an associative binary operation.

(vi) On \[{\bf{R}} - \left\{ { - 1} \right\}\]define \[a * b = \dfrac{a}{{b + 1}}\] ​
Now, we have to check for commutative.
∗ is commutative if \[a * b = b * a\]
Solving further, we
\[ \Rightarrow a * b = \dfrac{a}{{b + 1}}\]
\[ \Rightarrow b * a = \dfrac{b}{{a + 1}}\]
Since \[a * b \ne b * a\]
∗ is not commutative.
Now, we have to check for associations.
∗ is associative if \[(a * b) * c = a * (b * c)\]
\[ \Rightarrow (a * b) * c = \dfrac{a}{{b + 1}} * c = \dfrac{{\dfrac{a}{{b + 1}}}}{{c + 1}} = \dfrac{a}{{\left( {b + 1} \right)\left( {c + 1} \right)}}\]
 \[ \Rightarrow a * (b * c) = a * \dfrac{b}{{c + 1}} = \dfrac{a}{{\dfrac{b}{{c + 1}} + 1}} = \dfrac{{a(c + 1)}}{{b + c + 1}}\]
Since \[(a * b) * c \ne a * (b * c)\]
∗ is not an associative binary operation.

Note: We have to know the concept of binary operation. Additions and multiplications are the binary operations on each of the sets of Natural numbers, Integer, Rational number, Real Numbers, Complex number. Subtraction is a binary operation on each of the sets of Integer, Rational numbers, Real Numbers, Complex number. A division is not a binary operation on the set of Natural numbers, integer, Rational numbers, Real Numbers, Complex number. Exponential operation is a binary operation on the set of Natural numbers and not on the set of Integers. The additions and multiplication on the set of all irrational numbers are not the binary operations. Subtraction is not a binary operation on the set of Natural numbers.